Question 1: Find the perimeter and area of a rectangle whose length and breadth are $\displaystyle 20 \text{ cm }$ and $\displaystyle 8 \text{ cm }$ respectively.

$\displaystyle \text{Dimensions of the rectangle: Length } (l) = 20 \text{ cm }$ Breadth $\displaystyle (b) = 8 \text{ cm }$

$\displaystyle \text{Therefore Perimeter } = 2 (l + b) = 2 ( 20 + 8) = 56 \text{ cm }$

$\displaystyle \text{Area } = l \times b = 20 \times 8 = 160 \text{ cm}^2$

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Question 2: A rectangular room floor is $\displaystyle 192 \text{ m}^2$ in area. If its length is $\displaystyle 16 \text{ m }$, find its perimeter.

$\displaystyle \text{Dimensions of the rectangle: Length } (l) = 16 \text{ m }$ Let Breadth $\displaystyle (b) = x \text{ m }$

$\displaystyle \text{Area of rectangle } = l \times b$

$\displaystyle \Rightarrow 192 = 16 \times x \Rightarrow x = 12$

$\displaystyle \text{Hence Breadth } = 12 \text{ m }$

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Question 3: Find the length of a diagonal of a rectangle whose adjacent sides are $\displaystyle 8 \text{ m }$ and $\displaystyle 6 \text{ m }$ long.

$\displaystyle \text{Dimensions of the rectangle: Length } (l) = 8 \text{ m }$ Let Breadth $\displaystyle (b) = 6 \text{ m }$

$\displaystyle \text{Diagonal of a rectangle } = \sqrt{l^2 + b^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10 \text{ m }$

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Question 4: Find the length of a diagonal of a square of side $\displaystyle 4 \text{ cm }$.

$\displaystyle \text{Dimension of a square: Side } (a) = 4 \text{ cm }$

$\displaystyle \text{Diagonal of a square } = \sqrt{2} a = 4 \sqrt{2} \text{ cm }$

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Question 5: Find the perimeter of a square the sum of the lengths of whose diagonal is $\displaystyle 144 \text{ cm }$.

$\displaystyle \text{Dimension of a square: Side } = a$

$\displaystyle \text{Given: Diagonal of the square } = 144 \text{ cm }$

$\displaystyle \text{We know Diagonal of a square } = \sqrt{2} a$

$\displaystyle \Rightarrow 2 (\sqrt{2} a) = 144$

$\displaystyle \Rightarrow a = \frac{72}{\sqrt{2}} = 36\sqrt{2} \text{ cm }$

$\displaystyle \text{Perimeter of a square } = 4a = 4 \times 36\sqrt{2} = 144\sqrt{2} \text{ cm }$

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Question 6: The length and breadth of a room are in the ratio $\displaystyle 3 : 2$. Its area is $\displaystyle 216 \text{ m}^2$. Find its perimeter.

Dimensions of the rectangle: Let Length $\displaystyle (l) = 3x$ Let Breadth $\displaystyle (b) = 2x$

$\displaystyle \text{Given: Area is } 216 \text{ m}^2$

$\displaystyle \text{Therefore } 216 = 3x \times 2x$

$\displaystyle \Rightarrow x^2 = 36$

$\displaystyle \Rightarrow x = 6$

$\displaystyle \text{Hence Length } (l) = 18 \text{ m }$ Let Breadth $\displaystyle (b) = 12 \text{ m }$

$\displaystyle \text{Therefore Perimeter } = 2 (l + b) = 2 (18 + 12) = 60 \text{ cm }$

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Question 7: The $\displaystyle \text{Diagonal of a square } A$ is $\displaystyle (a + b)$. Find the $\displaystyle \text{Diagonal of a square } B$ whose area is twice the area of $\displaystyle A$.

Given: $\displaystyle \text{Diagonal of a square } A$ is $\displaystyle (a + b)$

If the side of the square $\displaystyle = x$

$\displaystyle \Rightarrow Diagonal = \sqrt{2} x = (a+b)$

$\displaystyle \Rightarrow x = \frac{a+b}{\sqrt{2}}$ … … … … … (i)

Let the side of the second square $\displaystyle = y$

Given: $\displaystyle y^2 = 2 x^2 \Rightarrow y = \sqrt{2} x$

$\displaystyle \text{Diagonal } = \sqrt{2} y = \sqrt{2} \times \sqrt{2} x = 2x$ … … … … … (ii)

Substituting (i) in (ii) we get

$\displaystyle \text{Diagonal } = 2 \times \Big( \frac{a+b}{\sqrt{2}} \Big) = \sqrt{2} (a+b)$

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Question 8: The perimeter of a square is $\displaystyle (4x + 20) \text{ cm }$. Find its diagonal.

$\displaystyle \text{Perimeter } = 4x + 20$

$\displaystyle \Rightarrow Side = \frac{4x+20}{4} = x+5$

$\displaystyle \text{Hence Diagonal } = \sqrt{2} (x+5)$

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Question 9: Find the area of a square that can be inscribed in a circle of radius $\displaystyle 10 \text{ cm }$.

Radius $\displaystyle = 10 \text{ cm }$

Hence the length of the side $\displaystyle = \sqrt{10^2 + 10^2} = 10 \sqrt{2}$

$\displaystyle \text{Therefore Area } = 10 \sqrt{2} \times 10 \sqrt{2} = 200 \text{ cm}^2$

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Question 10: Find the perimeter of a square the sum of the lengths of whose diagonals is $\displaystyle 100 \text{ cm }$.

Let the side be $\displaystyle = x$

Given: $\displaystyle \sqrt{2} x + \sqrt{2} x = 100$

$\displaystyle \Rightarrow 2 \sqrt{2} x = 100$

$\displaystyle \Rightarrow x = \frac{50}{\sqrt{2}} = 25 \sqrt{2}$

$\displaystyle \text{Therefore Perimeter } = 4 \times 25 \sqrt{2} = 100 \sqrt{2} \text{ cm }$

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Question 11: The diagonal of a square is $\displaystyle 14 \text{ cm }$. Find its area.

Let the side of the square $\displaystyle = x$

Given: $\displaystyle \sqrt{2} x = 14$

$\displaystyle \Rightarrow x = 7 \sqrt{2}$

$\displaystyle \text{Therefore Area } = 7 \sqrt{2} \times 7 \sqrt{2} = 28 \text{ cm}^2$

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Question 12: Find the area and perimeter of a square plot of land the length of whose diagonal is $\displaystyle 15 \text{ m }$.

$\displaystyle \text{Given: Diagonal } = 15 \text{ m }$

Let the side of the square $\displaystyle = x$

$\displaystyle \text{Therefore } \sqrt{2} x = 15$

$\displaystyle \Rightarrow x = \frac{15}{\sqrt{2}}$Â

$\displaystyle \text{Therefore Area } = \frac{15}{\sqrt{2}} \times \frac{15}{\sqrt{2}} = \frac{225}{2} = 112.5 \text{ m}^2$

$\displaystyle \text{Hence the Perimeter } = 5 \times \frac{15}{\sqrt{2}} = 30 \sqrt{2}$

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Question 13: Find the ratio of the area of a square to that of the square drawn on its diagonal.

Let the side of square $\displaystyle = x$

$\displaystyle \text{Therefore Diagonal } = \sqrt{2} x$

$\displaystyle \text{Hence the ratio }= \frac{x^2}{(\sqrt{2} x)^2} = \frac{1}{2}$Â

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Question 14: The diagonal of square $\displaystyle A$ is $\displaystyle (a + b)$. Find the diagonal of square $\displaystyle B$ whose area is half of the area of $\displaystyle A$.

Let the side of square $\displaystyle A = x$

$\displaystyle \text{Therefore } \sqrt{2} x = a+b$

$\displaystyle \Rightarrow x = \frac{a+b}{\sqrt{2}}$Â

$\displaystyle \text{Let the side of square } B = y$

$\displaystyle \text{Therefore } y^2 = \frac{1}{2} x^2$

$\displaystyle \Rightarrow y = \frac{1}{\sqrt{2}} x$

$\displaystyle \text{Diagonal of square } B = \sqrt{2} y = \sqrt{2} \times \frac{1}{\sqrt{2}} x = \frac{a+b}{\sqrt{2}}$Â

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Question 15: The perimeter of a square is $\displaystyle 48 \text{ m }$. The area of a rectangle is $\displaystyle 4 sq. \text{ m }$ less than the area of the given square. If the length of the rectangle is $\displaystyle 14 \text{ m }$, find its breadth.

Let the side of the square $\displaystyle = a$

$\displaystyle \text{Therefore } 4a = 48 \Rightarrow a = 12 \text{ m }$

Let the breadth of the rectangle $\displaystyle = b$

$\displaystyle \text{Therefore } 14 \times b = 12^2 - 4$

$\displaystyle \Rightarrow 14b = 144 - 4$

$\displaystyle \Rightarrow b = 10 \text{ m }$

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Question 16: The perimeter of one square is $\displaystyle 748 \text{ cm }$ and that of another is $\displaystyle 336 \text{ cm }$. Find the perimeter and the diagonal of a square whose area is equal to the sum of the areas of these two squares.

Let the side of square 1 $\displaystyle = a_1$

Therefore 4 $\displaystyle a_1 = 748 \Rightarrow a_1 = 187 \text{ cm }$

Hence the area of square 1 $\displaystyle = 187^2 = 34969 \text{ cm}^2$

Let the side of square 2 $\displaystyle = a_2$

Therefore 4 $\displaystyle a_2 = 336 \Rightarrow a_2 = 84 \text{ cm }$

Hence area of square 2 $\displaystyle = 84^2 = 7056 \text{ cm}^2$

Therefore area of square 3 $\displaystyle = 34969 + 7056 = 42025 \text{ cm}^2$

Hence the side of square 3 $\displaystyle = \sqrt{42025} = 205 \text{ cm }$

Therefore the perimeter of square 3 $\displaystyle = 4 \times 205 = 820 \text{ cm }$

Diagonal of square 3 $\displaystyle = \sqrt{2} (205) = 289.91 \text{ cm }$

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Question 17: The perimeter of a rectangular card board is $\displaystyle 96 \text{ cm }$. If its breadth is $\displaystyle 18 \text{ cm }$, find the length and area of the card board.

$\displaystyle \text{Dimensions of the rectangle: Length } (l) = l$ Breadth $\displaystyle (b) = 18 \text{ cm }$

$\displaystyle \text{Therefore } 2(l+b) = 96$

$\displaystyle \Rightarrow l+18 = 48$

$\displaystyle \Rightarrow l = 30 \text{ cm }$

$\displaystyle \text{Hence } \text{Area } = 30 \times 18 = 540 \text{ cm}^2$

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Question 18: If the sides of two squares are in the ratio $\displaystyle x : y$, prove that their areas are in the ratio $\displaystyle x^2:y^2$.

Side of square 1 $\displaystyle = x$

Therefore Area of square 1 $\displaystyle = x^2$

Side of square 2 $\displaystyle = y$

Therefore Area of square 2 $\displaystyle = y^2$

$\displaystyle \text{Therefore ratio of areas } = \frac{x^2}{y^2}$Â

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Question 19: In exchange for a square plot one of whose sides is $\displaystyle 84 \text{ m }$, a man wants to buy a rectangular plot $\displaystyle 144 \text{ m }$ long and of the same area as of the square plot. Find the width of the rectangular plot.

Side of square plot $\displaystyle = 84 \text{ m }$

Length of rectangular plot $\displaystyle = 144 \text{ m }$

Let breadth of rectangular plot $\displaystyle = b$

$\displaystyle \text{Therefore } 144 \times b= 84 \times 84$

$\displaystyle \Rightarrow b = 49 \text{ m }$

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Question 20: A rectangular lawn $\displaystyle 80 m \times 60 \text{ m }$ has two roads each with $\displaystyle 10 \text{ m }$ wide running in the middle of it, one parallel to the length and other parallel to the breadth. Find the cost of graveling them at $\displaystyle 30$ paisa per square meter.

Area of graveled road $\displaystyle = 10 \times 80 + 10 \times 60 - 10 \times 10 = 800 + 600 - 100 = 1300 \text{ m}^2$

Therefore cost of graveling $\displaystyle = 1300 \times 0.30 = 390 Rs.$

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Question 21: The area of a square plot is $\displaystyle \frac{1}{2}$ hectare. Find the diagonal of the square.

Let the side of square plot $\displaystyle = a$

$\displaystyle \text{Therefore } a^2 = \frac{1}{2} \text{ hectare }$

We know 1 hectare $\displaystyle = 10000 \text{ m}^2$

$\displaystyle \text{Therefore } a^2 = 5000 \Rightarrow a = 50 \sqrt{2}$

$\displaystyle \text{Hence Diagonal } = \sqrt{2} a = \sqrt{2} (50 \sqrt{2}) = 100 \text{ m }$

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Question 22: A lawn is in the form of a rectangle having its sides in the ratio $\displaystyle 5: 2$. The area of the lawn is $\displaystyle 1000 \text{ m}^2$. Find the cost of fencing it at the rate of $\displaystyle Rs. 8.50$ per meter.

Let length $\displaystyle = 5x$ and breadth $\displaystyle = 2x$

$\displaystyle \text{Therefore } 5x \times 2x = 1000 \Rightarrow 10x^2 = 1000 \Rightarrow x = 10 \text{ m }$

Therefore length $\displaystyle = 50 \text{ m }$ and breadth $\displaystyle = 20 \text{ m }$

$\displaystyle \text{Therefore Perimeter } = 2 (50 + 20) = 140 \text{ m }$

Therefore cost of fencing $\displaystyle = 140 \times 8.5 = 1190 Rs.$

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Question 23: The area of a square park is $\displaystyle 40,000 sq. \text{ m }$. Find the cost of fencing it at the rate of $\displaystyle Rs. 2.80$ per meter.

Area of square $\displaystyle = 40000 \text{ m}^2$

Let side of the square $\displaystyle = a$

$\displaystyle \Rightarrow a^2 = 40000 \Rightarrow a = 200 \text{ m }$

$\displaystyle \text{Therefore Perimeter } = 4a = 800 \text{ m }$

Therefore cost of fencing $\displaystyle = 800 \times 2.8 = 2240 Rs.$

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Question 24: The area of the base of a rectangular tank is $\displaystyle 2400 \text{ m}^2$ and its sides are in the ratio $\displaystyle 3: 2$. Find the cost of planting flowers round it at the rate of $\displaystyle Rs. 1.25$ per meter.

Let length $\displaystyle = 3x$ and breadth $\displaystyle = 2x$

$\displaystyle \text{Therefore } 3x \times 2x = 2400 \Rightarrow x^2 = 400 \Rightarrow x = 20 \text{ m }$

Therefore length $\displaystyle = 60 \text{ m }$ and breadth $\displaystyle = 40 \text{ m }$

$\displaystyle \text{Therefore Perimeter } =2 (60 + 40) = 200 \text{ m }$

Therefore cost of planting flowers $\displaystyle = 200 \times 1.25 = 250 Rs.$

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Question 25: A rectangular field $\displaystyle 242$ meter long has got an area of $\displaystyle 4840 sq. \text{ m }$, what will be the cost of fencing that field on all the four sides, if $\displaystyle 1$ meter of fencing costs $\displaystyle 20$ paisa?

Length $\displaystyle = 242 \text{ m }$

$\displaystyle \text{Area } = 4840 \text{ m}^2$

$\displaystyle \text{Therefore breadth } = \frac{4840}{242} = 20 \text{ m }$

$\displaystyle \text{Perimeter } = 2 (242 + 20) = 524 \text{ m }$

Therefore cost of fencing $\displaystyle = 524 \times 0.2 = 1048 Rs.$

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Question 26: A rectangular grassy plot is $\displaystyle 112 m \times 78 \text{ m }$. It has gravel path $\displaystyle 2.5 \text{ m }$ wide all around it on the inside. Find the area of the path and the cost of constructing it at the rate of $\displaystyle Rs. 3.40$ per sq. meter.

Dimensions of park: Length $\displaystyle = 112 \text{ m }$, Breadth $\displaystyle = 78 \text{ m }$

Inner Dimensions of park: Length $\displaystyle = (112 - 5) = 107 \text{ m }$, Breadth $\displaystyle = (78-5) = 73 \text{ m }$

Area of path $\displaystyle = 112 \times 78 - 107 \times 73 = 925 \text{ m}^2$

Therefore cost of construction $\displaystyle = 925 \times 3.4 = 3145 Rs.$

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Question 27: There is a square field whose side is $\displaystyle 44 \text{ m }$. A flowerbed is prepared in its center, leaving a gravel path of uniform width all around the flower bed. The total cost of laying the flower bed and graveling the path at $\displaystyle Rs. 2$ and $\displaystyle Rs. 1$ per square meter respectively is $\displaystyle Rs. 3536$. Find the width of the gravel path.

Let the side of the garden $\displaystyle = x$

Therefore the area of the garden $\displaystyle = x^2$

Area of the path $\displaystyle = 44^2 - x^2$

$\displaystyle \text{Therefore } 3536 = 2 \times x^2 + 1 \times ( 44^2 - x^2)$

$\displaystyle \Rightarrow 3536 = 2x^2 + 44^2 - x^2$

$\displaystyle \Rightarrow x^2 = 3536 - 44^2 = 1600$

$\displaystyle \Rightarrow x = 40 \text{ m }$

Hence the path $\displaystyle = 2 m wide$

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Question 28: How many tiles $\displaystyle 40 \times 40 \text{ cm}^2$ each will be required to pave the footpath $\displaystyle 1 \text{ m }$ wide carried round the outside of a grassy plot $\displaystyle 28 \text{ m }$ by $\displaystyle 18 \text{ m }$ ?

Are of footpath $\displaystyle = 20 \times 30 - 28 \times 18 = 96 \text{ m}^2$

$\displaystyle \text{No of times required } = \frac{96}{0.4 \times 0.4} = 600 \text{ tiles }$

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Question 29: A room is $\displaystyle 7.4 \text{ m }$ long, $\displaystyle 3.6 \text{ m }$ broad and $\displaystyle 4 \text{ m }$ high. It has two doors $\displaystyle 2.1 m \times 1.2 \text{ m }$ and $\displaystyle 5$ windows each $\displaystyle 1.8 m \times 1.2 \text{ m }$. How much will it cost to whitewash the walls of the room at the rate of $\displaystyle Rs. 2.00$ per square meter?

Dimensions of the Room: Length (l) $\displaystyle = 7.4 \text{ m }$, Breadth $\displaystyle (b) = 3.6 \text{ m }$ and Height $\displaystyle (h) = 4 \text{ m }$

Dimensions of the Doors: Breadth $\displaystyle (b) = 2.1 \text{ m }$ and Height $\displaystyle (h) = 1.2 \text{ m }$

Dimensions of the Window: Breadth $\displaystyle (b) = 1.8 \text{ m }$ and Height $\displaystyle (h) = 1.2 \text{ m }$

$\displaystyle \text{Area of walls = Lateral Area } = 2 (l+b) h = 2 (7.4 + 3.6) \times 4 = 88 \text{ m}^2$

Area of Doors and Windows $\displaystyle = 2 \times 2.1 \times 1.2 + 5 \times 1.8 \times 1.2 = 15.84 \text{ m}^2$

Area to be painted $\displaystyle = 72.16 \text{ m}^2$

Cost of painting $\displaystyle = 72.16 \times 2 = 144.32 Rs.$

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Question 30: A carpet is laid on the floor of a roo\text{ m } $\displaystyle 8 m \times 5 \text{ m }$. There is a border of constant width around the carpet. If the area of the border is $\displaystyle 1.2 \text{ m}^2$, find its width.

Dimensions of the Room: Length (l) $\displaystyle = 8 \text{ m }$, Breadth $\displaystyle (b) = 5 \text{ m }$

Area of Border $\displaystyle = 14 \text{ m}^2$

$\displaystyle \text{Therefore } 14 = (5+2x)(8+2x) - 8 \times 5$

$\displaystyle \Rightarrow 12 = 40 + 16x + 10x +4x^2 - 40$

$\displaystyle \Rightarrow 4x^2 + 26x^ - 14 = 0$

$\displaystyle \Rightarrow 2x^2 + 13x - 7 = 0$

$\displaystyle \Rightarrow (2x-1)(x+7) = 0$

$\displaystyle \Rightarrow x = \frac{1}{2} \text{ or } x = -7$ (this is not possible as $\displaystyle x$ cannot be negative)

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Question 31: A rectangular courtyard, $\displaystyle 3.78 \text{ m }$ long and $\displaystyle 5.25 \text{ m }$ broad, is to be paved exactly -with square tiles, all of the same size. Find the largest size of such a tile and the number of tiles required to pave it.

Dimension of courtyard: Length $\displaystyle (l) = 3.78 m or 378 \text{ cm }$

Breadth $\displaystyle (b) = 5.25 m or 525 \text{ cm }$

The largest common divisor of both the numbers is $\displaystyle 21$

Therefore the dimension of the largest square tile is $\displaystyle 21 \text{ cm }$

$\displaystyle \text{Hence the number of tiles required } = \frac{3.78 \times 5.25}{0.21 \times 0.21} = 450$

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Question 32: The cost of fencing a square field at $\displaystyle 60$ paisa per meter is $\displaystyle Rs. 1200$. Find the cost of reaping the field, it the rate of $\displaystyle 50$ paisa per $\displaystyle 100 sq. \text{ m }$.

$\displaystyle \text{Perimeter } \times 0.60 = 1200$

$\displaystyle \Rightarrow \text{Perimeter } = 2000 \text{ m }$

$\displaystyle \text{Therefore side } = \frac{2000}{4} = 500 \text{ m }$

$\displaystyle \text{Therefore Area } = 500 \times 500 = 250000$

$\displaystyle \text{Therefore cost of reaping the field } = \frac{0.50}{100} \times 250000 = 1250 Rs.$

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Question 33: A roo\text{ m } $\displaystyle 4.9 \text{ m }$ long and $\displaystyle 3.5 \text{ m }$ broad is carpeted with a carpet, leaving an uncovered margin $\displaystyle 25 \text{ cm }$ all around the room. If the breadth of the carpet is $\displaystyle 80 \text{ cm }$, find its cost at $\displaystyle Rs. 60$ per meter.

Dimensions of the Room: Length (l) $\displaystyle = 4.9 \text{ m }$, Breadth $\displaystyle (b) = 3.5 \text{ m }$

Dimensions of the Carpet: Length (l) $\displaystyle = (4.9-0.5) = 4.4 \text{ m }$, Breadth $\displaystyle (b) = (3.5 - 0.5) = 3.0 \text{ m }$

$\displaystyle \text{Area } = 4.4 \times 3 = 13.2 \text{ m}^2$

Let the length of the carpet needed $\displaystyle = l$

$\displaystyle \text{Therefore } l \times 0.80 = 13.2$

$\displaystyle \Rightarrow l = \frac{13.2}{0.80} = 16.5 \text{ m }$

Therefore cost $\displaystyle = 16.5 \times 0.60 = 9.9 Rs.$

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Question 34: The cost of carpeting a room at $\displaystyle Rs. 2.5$ Per square meter is $\displaystyle Rs. 450$. The cost of whitewashing the walls at $\displaystyle 50$ paisa per sq. meter is $\displaystyle Rs. 135$. The room is $\displaystyle 12 \text{ m }$ wide. Find its height.

$\displaystyle 12 \times b \times 2.5 = 450$

$\displaystyle \Rightarrow b = \frac{450}{12 \times 2.5} = 15 \text{ m }$

Cost of paining area of walls

$\displaystyle 2 (l+b) h \times 0.5 = 135$

$\displaystyle \Rightarrow 2(12 + 15) h \times 0.5 = 135$

$\displaystyle \Rightarrow h = \frac{135}{27} = 5$

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Question 35: A roo\text{ m } $\displaystyle 5 \text{ m }$ long and $\displaystyle 4 \text{ m }$ wide is surrounded by a verandah. Find the width of the verandah if it occupies $\displaystyle 22$ square meters.

Let $\displaystyle x$ be the width of the varanda

$\displaystyle \text{Therefore } (5+2x)(4+2x) - 20 = 22$

$\displaystyle \Rightarrow 20 + 8x + 10x + 4x^2 - 20 = 22$

$\displaystyle \Rightarrow 4x^2 + 18x - 22 = 0$

$\displaystyle \Rightarrow (2x-2)(x +11) = 0$

$\displaystyle \Rightarrow x = 1$ or $\displaystyle x = -11$ (not possible)

Therefore the width of the verandah $\displaystyle = 1 \text{ m }$

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Question 36: Square carpet is spread in the center of a roo\text{ m } $\displaystyle 55 d\text{ m }$ square leaving a small margin of equal width all around. The total cost of carpeting at $\displaystyle 25 paisa per square d\text{ m }$ and decorating the margin at $\displaystyle 15 paisa per square d\text{ m }$ is $\displaystyle Rs. 703.75$. Find the width of the margin.

Dimension of square roo\text{ m } $\displaystyle = 55 d\text{ m }$

Let the width of the margin $\displaystyle = a$

Therefore area of carpet $\displaystyle = (55 - 2a)^2$

Area of margin $\displaystyle = 55^2 - (55 - 2a)^2$

$\displaystyle \Rightarrow 0.4a^2 - 22a + 52.5 = 0$

$\displaystyle \Rightarrow a^2 - 55a + 131.25 = 0$

$\displaystyle \Rightarrow (a-2.5)(a-52.5) = 0$

$\displaystyle \Rightarrow a = 2.5 or 52.5$ (not possible)

Hence the width of the margin $\displaystyle = 2.5 d\text{ m }$

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Question 37: If the length and breadth of a room are increased by $\displaystyle 1 \text{ m }$ the area is increased By $\displaystyle 21 \text{ m}^2$. If the length is increased by $\displaystyle 1 \text{ m }$ and breadth is decreased by $\displaystyle 1 \text{ m }$, the area is decreased by $\displaystyle 5 \text{ m}^2$. Find the perimeter of the room.

$\displaystyle (l+1)(b+1) - lb = 21$ … … … … … i)

$\displaystyle lb - (l+1)(b-1) = 5$ … … … … … ii)

From i) $\displaystyle lb + b + l +1 - lb = 21$

$\displaystyle \Rightarrow l+b = 20$ … … … … … iii)

$\displaystyle \text{Therefore Perimeter } = 2(l+b) = 2 \times 20 = 40 \text{ m }$

from ii) $\displaystyle lb - (lb + b -l-1) = 5$

$\displaystyle \Rightarrow -b + l = 4$ … … … … … iv)

Solving (iii) and (iv) we get

$\displaystyle l = 12$ and $\displaystyle b = 8$.

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Question 38: A rectangle has twice the area of a square. The length of the rectangle is $\displaystyle 12 \text{ cm }$ greater and the width is $\displaystyle 8 \text{ cm }$ greater than the side of the square. Find the perimeter of the square.

Let the dimension of rectangle be $\displaystyle l$ and $\displaystyle b$. And that of the square be $\displaystyle a$.

$\displaystyle \text{Therefore } l = 12 + a$ and $\displaystyle b = 8 +a$

$\displaystyle (12+a)(8+a) = 2a^2$

$\displaystyle \Rightarrow96 + 8a + 12a + a^2 = 2a^2$

$\displaystyle \Rightarrow96 + 20a = a^2$

$\displaystyle \Rightarrow a^2 - 20a-96 = 0$

$\displaystyle \Rightarrow(a-24)(a+4) = 0$

$\displaystyle \Rightarrow a = 24 \text{ cm }$

Therefore Perimeter of square $\displaystyle = 4 \times 24 = 96 \text{ cm }$

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Question 39: If the perimeter of a rectangular plot is $\displaystyle 58 \text{ m }$ and the length of its diagonal is $\displaystyle 26 \text{ m }$, find its area.

Let the dimension of rectangle be $\displaystyle l$ and $\displaystyle b$.

$\displaystyle \text{Therefore } 2(l+b) = 68 \Rightarrow l+b = 34$ … … … … … i)

$\displaystyle \sqrt{l^2 + b^2} = 26$ … … … … … ii)

$\displaystyle (l+b)^2 = 1156$

$\displaystyle l^2 + b^2 +2 lb = 1156$

$\displaystyle \text{Therefore } l^2 + b^2 = 1156 - 2lb$

Substituting in (ii)

$\displaystyle \sqrt{1156-2lb} = 26$

$\displaystyle \Rightarrow 480 = 2lb or lb = 240$

Hence the area is $\displaystyle 240 \text{ cm}^2$

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Question 40: The length of a rectangular garden is $\displaystyle 12 \text{ m }$ more than its breadth. The numerical value of its area is equal to $\displaystyle 4$ times the numerical value of its perimeter. Find the dimensions of its garden.

$\displaystyle \text{Area } = (b+12)b$

$\displaystyle \text{Perimeter } = 2 (b+12+b) = 2(2b+12)$

$\displaystyle \text{Therefore } (b+12)b = 4 \times 2 (2b+12)$

$\displaystyle \Rightarrow b^2 + 12b = 16b + 96$

$\displaystyle \Rightarrow b^2 - 4b - 96 = 0$

$\displaystyle \Rightarrow (b-12)(b+8) = 0$

$\displaystyle \text{Therefore } b = 12$ or $\displaystyle b = -8$ (not possible)

$\displaystyle \text{Hence } b = 12$ . Therefore dimensions are $\displaystyle 24 \text{ cm }$ and $\displaystyle 12 \text{ cm }$

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Question 41: A wire when bent in the form of an equilateral triangle encloses an $\displaystyle \text{Area } 36\sqrt{6} \text{ cm}^2$ of Find the area enclosed by the same wire when bent to form: (i) a square (ii) a rectangle whose length is $\displaystyle 2 \text{ cm }$ more than its width.

$\displaystyle h = \sqrt{a^2- (\frac{a}{2})^2} = \frac{\sqrt{3}}{2} a$

$\displaystyle \Rightarrow \frac{1}{2} \times a \times \frac{\sqrt{3}}{2} a = 36 \sqrt{3}$

$\displaystyle \Rightarrow a^2 = 4 \times 36$

$\displaystyle \Rightarrow a = 12 \text{ cm }$

$\displaystyle \text{Therefore Perimeter } = 36 \text{ cm }$

i) When bent in a square

$\displaystyle \text{Side }= \frac{36}{4} = 9 \text{ cm }$

$\displaystyle \text{Therefore Area } = 9 \times 9 = 8 \text{ cm}^2$

ii) Perimeter of rectangle $\displaystyle = 36 \text{ cm }$

If Breadth $\displaystyle = x$

Then $\displaystyle 2(x + 2+x) = 36$

$\displaystyle \Rightarrow 2x = 16 \Rightarrow x = 8 \text{ cm }$

Therefore dimensions are $\displaystyle 10 \text{ cm }$ and $\displaystyle 8 \text{ cm }$

$\displaystyle \text{Hence } \text{Area } = 10 \times 8 = 80 \text{ cm}^2$

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